How do you solve the following linear system: #8x+3y-6z=4, x-2y-z=2, 4x+y-2z=-4 #?

1 Answer
Dec 8, 2015

Answer:

Combine #2# pairs of equations to eliminate 1 of the variables from 2 equations; then solve the 2 equations for each of the remaining variables.
#color(white)("XXX")(x,y,z)=(-4,0,-6)#

Explanation:

Given:
[1]#color(white)("XXX")8x+3y-6z=4#
[2]#color(white)("XXX")x-2y-z=2#
[3]#color(white)("XXX")4x+y-2z=-4#

To eliminate #x# from 2 equations:

Multiply [2] by #8#
[4]#color(white)("XXX")8x-16y-8z=16#
Subtract [4] from [1]
[5]#color(white)("XXX")19y+2z=-12#

Multiply [3] by #2#
[6]#color(white)("XXX")8x+2y-4z=-8#
Subtract [6] from [1]
[7]#color(white)("XXX")y-2z=12#

Now we have two equations in two unknowns:
[5]#color(white)("XXX")19y+2z=-12#
[7]#color(white)("XXX")y-2z=12#

Adding [5] and [7]
[8]#color(white)("XXX")20y = 0#
[9]#color(white)("XXX")y=0#

Substituting #0# for #y# in [7] (could have used [5])
[10]#color(white)("XXX")(0)-2z=12#
[11]#color(white)("XXX")z=-6#

Substituting #(-6)# for #z# and #0# for #y# in [2] (could have used [1] or [3])
[12]#color(white)("XXX")x-2*(0)-(-6) = 2#
[13]#color(white)("XXX")x=-4#